Let the normal at a point $P$ on the curve $y^{2}-3 x^{2}+y+10$ $=0$ intersect the $y$-axis at $\lef

Question

Let the normal at a point $P$ on the curve $y^{2}-3 x^{2}+y+10$ $=0$ intersect the $y$-axis at $\left(0, \frac{3}{2}\right)$. If $m$ is the slope of the tangent at $P$ to the curve, then $|m|$ is equal to

Professor PC Thomas Classes · Answer

(4) : Let $P \equiv\left(x_{1}, y_{1}ight)$ be any point on the curve $y^{2}-3 x^{2}+y+10=0$\ Now, $2 y y^{\prime}-6 x+y^{\prime}=0 \Rightarrow y^{\prime}=\left(\frac{6 x_{1}}{1+2 y_{1}}ight)$\ Slope of normal $=\frac{-1}{	ext { slope of tangent }}$\ $\Rightarrow\left(\frac{\frac{3}{2}-y_{1}}{-x_{1}}ight)=-\left(\frac{1+2 y_{1}}{6 x_{1}}ight)$\ $\Rightarrow \quad 9-6 y_{1}=1+2 y_{1} \Rightarrow y_{1}=1$\ $	herefore \quad x_{1}= \pm 2$\ $	herefore$ Slope of tangent $=\frac{ \pm 12}{3}= \pm 4$. So, $|m|=4$\